- A square contains 16 cells. Divide the square into two equal parts so that the cut line goes along the sides of the cells. (Methods of cutting a square into two parts will be considered different if the parts of the square obtained by one method of cutting are not equal to the parts obtained by another method.) How many total solutions does the problem have?
- A 3X4 rectangle contains 12 cells. Find five ways to cut a rectangle into two equal parts so that the cut line goes along the sides of the cells (cutting methods are considered different if the parts obtained by one cutting method are not equal to the parts obtained by another method).
- A 3X5 rectangle contains 15 cells and the central cell has been removed. Find five ways to cut the remaining figure into two equal parts so that the cutting line goes along the sides of the cells.
- A 6x6 square is divided into 36 identical squares. Find five ways to cut a square into two equal parts so that the cutting line goes along the sides of the squares. Note: the problem has more than 200 solutions.
- Divide the 4x4 square into four equal parts, with the cut line running along the sides of the squares. How many different cutting methods can you find?
- Divide the figure (Fig. 5) into three equal parts so that the cut line runs along the sides of the squares.
7. Divide the figure (Fig. 6) into four equal parts so that the cut line runs along the sides of the squares.
8. Divide the figure (Fig. 7) into four equal parts so that the cut lines go along the sides of the squares. Find as many solutions as possible.
9. Divide the 5x5 square with the center square cut out into four equal parts.
10. Cut the figures shown in Fig. 8 into two equal parts along the grid lines, and each part should have a circle.
11. The figures shown in Fig. 9 must be cut along the grid lines into four equal parts so that each part has a circle. How to do it?
12. Cut the figure shown in Fig. 10 along the grid lines into four equal parts and fold them into a square so that the circles and stars are located symmetrically with respect to all axes of symmetry of the square.
13. Cut this square (Fig. 11) along the sides of the cells so that all parts are the same size and shape and so that each contains one circle and an asterisk.
14. Cut the 6x6 checkered paper square shown in Figure 12 into four equal pieces so that each piece contains three shaded squares.
7th grade club
Head Varvara Alekseevna Kosorotova
2009/2010 academic year
Lesson 8. Cutting on a checkered sheet of paper
When solving problems of this type, it is useful to apply the following considerations:
- Square. If you need to split a figure into several equal parts, you should first find the area of the figure being cut, and then find the area of each of the parts. Similarly, if the original figure needs to be divided into several figures of a given type, it is worth first calculating how many there should be. The same considerations can help when solving other cutting problems. To illustrate this idea, the author of these lines added problem 13 to the list, which was not among the problems offered in the lesson.
- Symmetry. Attention should be paid to the properties of symmetry, for example, in the case when it is necessary to cut one figure into parts and assemble another figure from them.
Divide the 4x4 square into two equal parts in four different ways, so that the cut line runs along the sides of the squares. Flag - 1. Cut the 6-stripe flag into two pieces so that you can fold them into an 8-stripe flag.
Flag - 2. Cut flag A into four pieces so that flag B can be folded from them. 
Cut the figure into 4 equal parts. 
Of the two - one. Cut the square with the hole in two straight lines into 4 pieces so that you can fold a new square from them and another regular 5x5 square. 
11*.
Jagged square. Turn a jagged square into a regular square by cutting it into 5 pieces. 
12*.
Maltese cross - 2. Cut the “Maltese cross” (see problem 8) into 5 pieces so that they can be folded into a square. 13**. Dunno cut the figure shown in the figure into three-cell and four-cell corners (such as in the picture). How many corners could Dunno get? Consider all possible cases! 
Solution. The area of the original figure is 22 (we take one cell as a unit of area). Let n four-cell and k three-cell corners be used for cutting. Then we express the area of the large figure as the sum of the areas of the corners: 22 = 3 k + 4 n. Let's rewrite this equality in this form: 22 − 4 n =3 k. On the left side of this equality there is an even number, which, however, is not divisible by 4. This means that 3 k is also an even number, not divisible by 4, and therefore the number k itself is such. In addition, on the right side of the equality there is a number that is a multiple of 3, so 22 − 4 n is also a multiple of 3. Thus, 22 − 4 n is a multiple of 6. Going through the values of n from 0 to 5 (for n ≥6 22 − 4 n<0<3 k , чего быть не может), получаем, что такое возможно лишь при n =1 и при n =4. В каждом из этих случаев несложно найти k . При n =1 имеем k =6, а при n =4 имеем k =2.
Note that we have not yet proven that both of these cases are realized. After all, equality of areas is only a necessary condition for the existence of a cutting method, but in no way sufficient (for example, a rectangle of size 1 × 6, obviously, cannot be cut into two three-cell corners, although 3 2 = 6). To complete the proof, examples of cuts of each type should be given. This can be done in many different ways. The picture shows only one of them, and you can try to come up with something of your own. By the way, it would be interesting to answer this question: how many cuts of each type are there? (The author of these lines, for example, does not yet know the answer to this question). 
In conclusion, we emphasize once again that a complete solution to this problem involves two steps: finding possible cases and checking that all of them are realized. Each of these steps alone is not a solution to the problem!
Teacher's opening remarks:
A little historical background: Many scientists have been interested in cutting problems since ancient times. Solutions to many simple cutting problems were found by the ancient Greeks and Chinese, but the first systematic treatise on this topic was written by Abul-Vef. Geometers began seriously solving problems of cutting figures into the smallest number of parts and then constructing another figure in the early 20th century. One of the founders of this section was the famous puzzle founder Henry E. Dudeney.
Nowadays, puzzle lovers are keen on solving cutting problems because there is no universal method for solving such problems, and everyone who undertakes to solve them can fully demonstrate their ingenuity, intuition and ability for creative thinking. (During the lesson we will indicate only one of the possible examples of cutting. It can be assumed that students may end up with some other correct combination - there is no need to be afraid of this).
This lesson is supposed to be conducted in the form of a practical lesson. Divide the circle participants into groups of 2-3 people. Provide each group with figures prepared in advance by the teacher. Students have a ruler (with divisions), a pencil, and scissors. It is allowed to make only straight cuts using scissors. Having cut a figure into pieces, you need to make another figure from the same parts.
Cutting tasks:
1). Try cutting the figure shown in the figure into 3 equal-shaped parts:
Hint: The small shapes look a lot like the letter T.
2). Now cut this figure into 4 equal-shaped parts:
Hint: It is easy to guess that small figures will consist of 3 cells, but there are not many figures with three cells. There are only two types: corner and rectangle.
3). Divide the figure into two equal parts, and use the resulting parts to form a chessboard.
Hint: Suggest starting the task from the second part, as if getting a chessboard. Remember what shape a chessboard has (square). Count the available number of cells in length and width. (Remember that there should be 8 cells).
4). Try cutting the cheese into eight equal pieces with three movements of the knife.
Tip: try cutting the cheese lengthwise.
Tasks for independent solution:
1). Cut out a square of paper and do the following:
· cut into 4 pieces that can be used to make two equal smaller squares.
· cut into five parts - four isosceles triangles and one square - and fold them so that you get three squares.
Cutting problems are an area of mathematics where, as they say, there are no mammoths lying around. Many individual problems, but essentially no general theory. Apart from the well-known Bolyai-Gerwin theorem, there are practically no other fundamental results in this area. Uncertainty is an eternal companion to cutting tasks. We can, for example, cut a regular pentagon into six pieces, from which we can form a square; however, we cannot prove that five parts would not be enough for this.
With the help of cunning heuristics, imagination and half a liter, we sometimes manage to find a specific solution, but, as a rule, we do not have the appropriate tools to prove the minimality of this solution or its non-existence (the latter, of course, applies to the case when we have not found a solution) . It's sad and unfair. And one day I took a blank notebook and decided to restore justice on the scale of one specific task: cutting a flat figure into two equal (congruent) parts. As part of this series of articles (by the way, there will be three of them), you and I, comrades, will look at this funny polygon shown below and try to impartially figure out whether it is possible to cut it into two equal figures or not.
Introduction
First, let's refresh our school geometry course and remember what equal figures are. Yandex helpfully suggests:Two figures on a plane are called equal if there is a movement that one-to-one transforms one figure into the other.
Now let's ask Wikipedia about movements. She will tell us, firstly, that motion is a transformation of the plane that preserves the distances between points. Secondly, there is even a classification of movements on a plane. They all belong to one of the following three types:
- Gliding symmetry (here, for the sake of convenience and benefit, I include mirror symmetry, as a degenerate case, where parallel translation is carried out to the zero vector)
Let us introduce some notation. We will call the figure being cut figure A, and the two hypothetical equal figures into which we supposedly can cut it will be called B and C, respectively. We will call the part of the plane not occupied by figure A region D. In cases where a specific polygon from the picture is considered as the cut figure, we will call it A 0 .
So, if figure A can be cut into two equal parts B and C, then there is a movement that translates B into C. This movement can be either parallel translation, or rotation, or sliding symmetry (from now on, I no longer stipulate that mirror symmetry is also considered sliding). Our decision will be built on this simple and, I would even say, obvious basis. In this part we will look at the simplest case - parallel transfer. Rotation and sliding symmetry will fall into the second and third parts respectively.
Case 1: parallel transfer
Parallel transfer is specified by a single parameter - the vector by which the shift occurs. Let's introduce a few more terms. A straight line parallel to the shift vector and containing at least one point of the figure A will be called secant. The intersection of a secant line and figure A will be called cross section. A secant with respect to which the figure A (minus the section) lies entirely in one half-plane will be called border.
Lemma 1. A boundary section must contain more than one point.
Proof: obvious. Well, or in more detail: let’s prove it by contradiction. If this point belongs to figure B, then it image(i.e., the point to which it will go during parallel translation) belongs to figure C => the image belongs to figure A => the image belongs to the section. Contradiction. If this point belongs to figure C, then it prototype(the point that, with parallel translation, will go into it) belongs to figure B, and then similarly. It turns out that there must be at least two points in the section.
Guided by this simple lemma, it is not difficult to understand that the desired parallel translation can only occur along the vertical axis (in the current orientation of the picture). If it were in any other direction, at least one of the boundary sections would consist of a single point. This can be understood by mentally rotating the shift vector and seeing what happens to the boundaries. To eliminate the case of vertical parallel transfer, we need a more sophisticated tool.
Lemma 2. The inverse image of a point located on the boundary of figure C is either on the boundary of figures B and C, or on the boundary of figure B and region D.
Proof: not obvious, but we'll fix it now. Let me remind you that the boundary point of a figure is such a point that, no matter how close to it, there are both points that belong to the figure and points that do not belong to it. Accordingly, near the boundary point (let's call it O") of figure C there will be both points of figure C and other points belonging to either figure B or region D. The inverse images of points of figure C can only be points of figure B. Consequently, arbitrarily close to the inverse image of the point O" (it would be logical to call it point O) there are points of the figure B. The inverse images of the points of the figure B can be any points that do not belong to B (that is, either the points of the figure C or the points of the region D). Similarly for points of region D. Consequently, no matter how close to point O there are either points of figure C (and then point O will be on the boundary of B and C) or points of region D (and then the inverse image will be on the boundary of B and D). If you can get through all these letters, you will agree that the lemma is proven.
Theorem 1. If the cross-section of figure A is a segment, then its length is a multiple of the length of the shift vector.
Proof: consider the “far” end of this segment (i.e., the end whose prototype also belongs to the segment). This end obviously belongs to figure C and is its boundary point. Consequently, its inverse image (by the way, also lying on the segment and separated from the image by the length of the shift vector) will be either on the boundary of B and C, or on the boundary of B and D. If it is on the boundary of B and C, then we also take its inverse image . We will repeat this operation until the next inverse image ceases to be on the boundary C and ends up on the boundary D - and this will happen exactly at the other end of the section. As a result, we get a chain of preimages that divide the section into a number of small segments, the length of each of which is equal to the length of the shift vector. Therefore, the length of the section is a multiple of the length of the shift vector, etc.
Corollary to Theorem 1. Any two sections that are segments must be commensurate.
Using this corollary, it is easy to show that vertical parallel transfer also disappears.
Indeed, section one has a length of three cells, and section two has a length of three minus the root of two in half. Obviously, these values are incommensurable.
Conclusion
If figure A 0 and can be cut into two equal figures B and C, then B is not translated into C by parallel translation. To be continued.With a sheet of checkered paper using scissors you can solve many different and interesting problems. These tasks are not only interesting or fun. They often contain a practical solution and proof of sometimes very complex geometric questions.
Let's start with the main rule of cutting and folding: Two polygons are called equicomposite if one of them can be divided (cut) into some other polygons, from which the second polygon can then be formed.
Equally proportioned polygons, of course, have the same area (equal in size), and therefore the property of equicomposition sometimes allows us to obtain formulas for calculating areas or compare the areas of figures (as they say, method of partitioning or decomposition). An example is comparing (calculating) the areas of a parallelogram and a rectangle.
The general question of the equivalence of two polygons is far from simple. There is an amazing theorem that states that from any given polygon, by cutting it into pieces, any other polygon of the same area can be constructed.
This theorem deals with so-called simple polygons. A simple polygon is a polygon whose boundary consists of one closed line without self-intersections, and exactly two of its links converge at each vertex of this broken line. An important property of a simple polygon is the fact that it has at least one internal diagonal.
Note that in order to allow the transformation of a rectangle into a square, we (Figure 3) needed to divide it into three parts. However, this partition is not the only one. For example, you can give an example of dividing a rectangle into four parts (Figure 4).
https://pandia.ru/text/78/456/images/image005_116.gif" width="356" height="391 src=">
The question of what smallest number of cuts is enough to construct another from one figure remains open to this day.
Task 1.
One woman had a rectangular rug measuring 27 by 36 inches; two opposite corners were frayed (Figure 5) and had to be cut off, but she wanted a rectangular rug. She gave this job to the master and he did it. How did he do this?
The solution to the problem can be seen from Figure 6.
https://pandia.ru/text/78/456/images/image009_72.gif" width="286" height="240 src=">
If toothed part A is removed from toothed part B and then pushed back between the teeth of part B, moving it one tooth to the right, the desired rectangle will be obtained.
Task 2.
How to make a square from five identical squares by cutting them.
As shown in Figure 7, four squares need to be cut into a triangle and a trapezoid. Attach four trapezoids to the sides of the fifth square and, finally, attach the triangles with their legs to the bases of the trapezoids.
https://pandia.ru/text/78/456/images/image011_68.gif" width="382" height="271 src=">
Task 3.
Cut the square into seven such pieces so that, when you add them, you get three equal squares. (Figures 8, 9)
https://pandia.ru/text/78/456/images/image013_60.gif" width="188" height="189 src="> 
Task 4.
Cut the square into eight pieces so that by adding them you get two squares, one of which is half the size of the other.
From Figure 10 you can see how to cut the square. The solution is similar to the solution to the previous problem. Figure 11 shows how to add the pieces to get the two required squares.
Educational tour
Tasks for teams of the “younger” age group to independently solve
Problem 1
A snail crawls up a pole 10 m high. During the day it rises 5 m, and during the night it drops 4 m. How long will it take the snail to get from the bottom to the top of the pole?
Problem 2
Is it possible to cut a hole in a piece of notebook paper that a person could fit through?
Problem 3
Hares are sawing a log. They made 10 cuts. How many logs did you get?
Problem 4
The bagel is cut into sectors. We made 10 cuts. How many pieces did you get?
Problem 5
On a large round cake, 10 cuts were made so that each cut goes from edge to edge and passes through the center of the cake. How many pieces did you get?
Problem 6
Two people had two square cakes. Everyone made 2 straight cuts on their cake from edge to edge. At the same time, one got three pieces, and the other got four. How could this be?
Problem 7
The hares are sawing the log again, but now both ends of the log are secured. Ten middle logs fell, but the two outer ones remained fixed. How many cuts did the hares make?
Problem 8
How to divide a pancake into 4.5, 6, 7 parts using three straight cuts?
Problem 9
There is a round chocolate bar on a rectangular cake. How to cut a cake into two equal parts so that the chocolate bar also splits exactly in half?
Problem 10
Is it possible to bake a cake that can be divided into 4 parts with one straight cut?
Problem 11
What is the maximum number of pieces that a round pancake can be divided into using three straight cuts?
Problem 12
How many times longer is the staircase to the fourth floor of a house than the staircase to the second floor of the same house?
Problem 13
Giuseppe has a sheet of plywood, size 22 × 15. Giuseppe wants to cut out as many rectangular pieces of size 3 from it as possible. × 5. How to do this?
Problem 14
The Magic Land has its own magical laws of nature, one of which says: “A flying carpet will fly only when it has a rectangular shape.”
Ivan Tsarevich had a magic carpet size 9 × 12. One day the Serpent Gorynych crept up and cut off a small rug of size 1 from this carpet × 8. Ivan Tsarevich was very upset and wanted to cut off another piece 1 × 4 to make a rectangle 8 × 12, but Vasilisa the Wise suggested doing differently. She cut the carpet into three parts, from which she used magic threads to sew a square flying carpet measuring 10 × 10.
Can you guess how Vasilisa the Wise remade the damaged carpet?
Problem 15
When Gulliver got to Lilliput, he discovered that all things there were exactly 12 times shorter than in his homeland. Can you tell how many Lilliputian matchboxes will fit in Gulliver's matchbox?
Problem 16
A two-color rectangular flag flutters from the mast of a pirate ship, consisting of alternating black and white vertical stripes of the same width. The total number of stripes is equal to the number of prisoners currently on the ship. At first there were 12 prisoners on the ship, and 12 stripes on the flag; the two prisoners then escaped. How to cut a flag into two parts and then sew them together so that the area of the flag and the width of the stripes do not change, but the number of stripes becomes 10?
Problem 17
A point was marked in the circle. Is it possible to cut this circle into three parts so that they can be used to form a new circle, with the marked point in the center?
Problem 18
Is it possible to cut a square into four parts so that each part touches (i.e. has common areas of the border) with the other three?
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Problem 24
There are no divisions on a 9 cm long ruler. Apply three intermediate divisions on it so that you can use it to measure distances from 1 to 9 cm with an accuracy of 1 cm.
Problem 25
Write some numbers near each vertex of the triangle, and write the sum of the numbers at the ends of that side near each side of the triangle. Now add each number near the top to the number near the opposite side. Why do you think the amounts turned out to be the same?
Problem 26
What is the area of a triangle with sides 18, 17, 35?
Problem 27
Cut the square into five triangles so that the area of one of these triangles is equal to the sum of the areas of the remaining ones.
Problem 28
A square sheet of paper was cut into six pieces in the shape of convex polygons; five pieces were lost, leaving one piece in the shape of a regular octagon (see picture). Is it possible to reconstruct the original square using this octagon alone?
Problem 29
You can easily cut a square into two equal triangles or two equal quadrangles. How to cut a square into two equal pentagons or two equal hexagons?
Problem 30
Ivan Tsarevich went to look for Vasilisa the Beautiful, who had been kidnapped by Koshchei. Leshy meets him.
“I know,” he says, “I used to go there and go to the Kingdom of Koshcheevo.” I walked for four days and four nights. In the first 24 hours I walked a third of the way, the straight road to the north. Then he turned west, trudged through the forest for a day, and walked half as far. For the third day I walked through the forest, already to the south, and came out on a straight road leading to the east. I walked along it 100 miles in a day and ended up in the Koshcheevo kingdom. You are as fast a walker as I am. Go, Ivan Tsarevich, look, on the fifth day you will be visiting Koshchei.
No,” answered Ivan Tsarevich, “if everything is as you say, then tomorrow I will see my Vasilisa the Beautiful.”
Is he right? How many miles did Leshy walk and how far does Tsarevich Ivan think about walking?
Problem 31
Come up with a color scheme for the faces of the cube so that in three different positions it looks like the one shown in the picture. (Specify how to color invisible edges, or draw a net.)
https://pandia.ru/text/78/456/images/image023_44.gif" align="left" width="205" height="205 src="> Problem 32
Numismatist Fedya has all coins with a diameter of no more than 10 cm. He stores them in a flat box measuring 30 cm * 70 cm (in one layer). He was given a coin with a diameter of 25 cm. Prove that all the coins can be placed in one flat box measuring 55 cm * 55 cm.
Problem 33
A central square was cut out of a 5x5 square. Cut the resulting shape into two parts into which you can wrap a 2x2x2 cube.
Problem 34
Cut this square along the sides of the cells into four parts so that all parts are the same size and the same shape and so that each part contains one circle and one star.
Problem 35
The parking lot in the Flower City is a square of 7x 7 cells, in each of which you can park a car. The parking lot is surrounded by a fence, one of the sides of the corner cage is removed (this is the gate). The car drives along a cage-wide path. Dunno was asked to place as many cars as possible in the parking lot so that any one could leave when others were standing. Dunno arranged 24 cars as shown in Fig. Try arranging the cars differently to accommodate more of them.
Problem 36
Petya and Vasya live in neighboring houses (see plan in the picture). Vasya lives in the fourth entrance. It is known that Petya, in order to reach Vasya by the shortest route (not necessarily going along the sides of the cells), does not care which side he runs around his house. Determine which entrance Petya lives in.
Problem 37
Suggest a way to measure the diagonal of an ordinary brick, which is easily implemented in practice (without the Pythagorean theorem).
Problem 38
Cut a cross made of five identical squares into three polygons equal in area and perimeter.
Problem 39
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Problem 46
a) Tetrahedron b) the cube was cut along the edges highlighted with bold lines (see pictures) and unfolded. Draw the resulting developments.
Problem 47
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